Solve FizzBuzz in Elixir

FizzBuzz with multiple solutions in Elixir. Which is best?

Why FizzBuzz?

It’s a problem everyone knows, so we can focus more on the way I solve it, than how I solve it.

The Solutions

Elixir is a functional language. And it’s very flexible.

This means there are tons of ways to solve a problem like FizzBuzz, and sometimes that means too many.

Let’s take a look at two ways we can use Elixir’s features to solve FizzBuzz in a clean and readable way. I don’t think one is better than the other, that’s for you to decide.

Though, it’s my opinion that the one that doesn’t make you say ugh, or scratch your head, is the better one.

“Whenever I have to think to understand what the code is doing, I ask myself if I can refactor the code to make that understanding more immediately apparent.” — Martin Fowler

1. Guards

https://hexdocs.pm/elixir/guards.html

The first solution uses Elixir’s guards. A feature that augments pattern matching by allowing you to add some additional checks to your function. i.e. ( from the docs)

def empty_map?(map) when map_size(map) == 0, do: true
def empty_map?(map) when is_map(map), do: false

Here’s what is looks like when applied to FizzBuzz

defmodule GaurdFizzBuzz do
  def solve(min, max), do: Enum.each(min..max, &(solve(&1)))
  def solve(num) when (rem(num, 15) == 0), do: IO.puts "FizzBuzz"
  def solve(num) when (rem(num, 5) == 0), do: IO.puts "Buzz"
  def solve(num) when (rem(num, 3) == 0), do: IO.puts "Fizz"
  def solve(num), do: IO.puts num
end

2. Conditionals

https://elixir-lang.org/getting-started/case-cond-and-if.html

The second solution uses Elixir’s conditionals. A way to match based off of some operations. Like a condensed and simple way to do if-else's

They look a bit like switch statements, but operate like an if clause.

i.e. ( from the docs)

cond do
  2 + 2 == 5 -> "This is never true"
  2 * 2 == 3 -> "Nor this"
  true -> "This is always true (equivalent to else)
end

Here’s what is looks like when applied to FizzBuzz!

defmodule MatchFizzBuzz do
  def solve(min, max), do: Enum.each(min..max, &(solve(&1)))
  def solve(num) do
    cond do
      rem(num, 15) == 0 -> IO.puts "FizzBuzz"
      rem(num, 5) == 0  -> IO.puts "Buzz"
      rem(num, 3) == 0  -> IO.puts "Fizz"
      true              -> IO.puts num
    end
  end
end

Conclusion

Which is better? Each has its merits. And as far as which I would rather see in a production code-base. I’m not really certain. Let’s take a look at the pros and cons:

Gaurds 1. Are not as readable when used for lots of pattern matching 2. Would be more extendable and testable.

Conds 1. Are more readable. 2. Would be harder to extend if some other requirements came down. 3. You’d have to refactor it if you wanted to add additional logic to it.

It’s really up to you and your team which you’d rather work-with. Often the finally level of optimization (guard v.s. cond) is not worth it. Shipped code is the best code.

So there you have it! FizzBuzz solved in two ways, each using a different language feature of Elixir.

Thanks for reading :) Catch me on twitter https://twitter.com/jwbaldwin_